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In Mathematics, a curve that is obtained by the intersection of a cone’s surface with a plane is known as a conic section. It is defined through a second-degree polynomial equation. Conic sections are of four types: Circle, Ellipse, Parabola, and Hyperbola.
Each of these types is obtained by a cross-section of a plane cutting through a cone. These types of curves are represented or defined through the help of different terms such as Latus rectum, focus, directrix, etc. In this article, we will discuss the latus rectum of an ellipse.
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What is a Latus rectum?
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The term latus rectum is actually a combination of Latin words wherein “Latus” means side and “Rectum” means straight. It can also be defined as the chord passing through the focus and perpendicular to the directrix. The endpoint of the Latus Rectum lies on its perimeter i.e. on its curve. The Half of the Latus Rectum is known as the Semi Latus Rectum.
Discover about the Chapter video:
Conic Sections Detailed Video Explanation:
Ellipse
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An ellipse is defined as a plane curve that surrounds two focal points. It does so in a way that for all points on the curve, the sum of two distances from some fixed point to the focal point remains a constant.
- The two fixed points are called Foci of the ellipse.
- The centre of the line segment that joins the two foci of the ellipse is called the Centre of the ellipse.
- The line joining the two foci is known as the major axis.
- The line perpendicular to the major axis is called the minor axis.
- The ending points of the major axis, touching the ellipse are called the Vertices of the ellipse.
In general representations,
- In the ellipse, the length of the major axis is denoted by 2a.
- In the ellipse, the length of the minor axis is denoted by 2b.
- In the ellipse, the distance between the foci is denoted by 2c.
Latus Rectum of Ellipse
The latus rectum of an ellipse is defined as the length of the line segment perpendicular to the major axis, passing through any of the foci, whose endpoints lie on the ellipse.
The formula for finding the length of the latus rectum of an ellipse is
2b2/a
Let us understand it further by demonstrating the concepts through some examples-
Example of Latus rectum of Ellipse
Find the equation of the latus rectum of an ellipse that is represented by the following equation:
9x2 + 4y2 – 18 x − 8 y − 23 = 0
Answer: 9x2 + 4y2 − 18x − 8y − 23=0
⇒ (3x−3)2 + (2y−2)2 −13−23=0
⇒ 9(x−1)2 + 4(y−1)2 = 36
⇒ 4(x−1)2 + 9(y−1)2 = 1
Shifting origin to (1, 1)
⇒ x−1 = x, y−1 = y
⇒ 4x2 + 9y2 = 1
a=2, b=3, e2 = 1+ (b2) *( a2) = 1 + 94 = 95
⇒ e = ±√59+ √53
⇒ Focus = (0, ±be) = (0, ±√5)
⇒ Latus ractum ⇒ y = ±√5
Shifting back, y = y−1
⇒ y − 1 = ±√5
⇒ y = 1 ± √5.
Sample Questions of Latus rectum of Ellipse
Question: Find the length of the latus rectum of an ellipse 4x2 + 9y2 – 24x + 36y – 72 = 0. (3 marks)
Solution:
We are given that the equation of the ellipse is 4x2 + 9y2 – 24x + 36y – 72 =0
⇒ (4x2 – 24x) + (9y2 + 36y) – 72 = 0
⇒ 4(x2 -6x) + 9(y2 + 4y) – 72 = 0
⇒ 4[x2 – 6x +9] + 9[y2 + 4y +4] = 144
⇒ 4(x – 3)2 + 9(y + 2)2 = 144
⇒ {(x – 3)2/36} + {(y + 2)2/16 } = 1
⇒ {(x– 3)2/62} + {(y + 2)2/42 } = 1
⇒ a=3 and b=2
Therefore, the length of the latus rectum for the given ellipse is:
= 2b2/a
= 2(2)2 /3
= 2(4)/3
= 8/3
The length of the latus rectum of the given ellipse 4x2 + 9y2 – 24x + 36y – 72 = 0 is 8/3.
Question: Let the length of the latus rectum of an ellipse, having its centre at the origin and the major axis along the x-axis, be 8. If the distance between the foci is equal to the length of the minor axis of the ellipse, then which of the following points lies on it? (4 marks)
(42, 2√2)
(43, 2√2)
(43, 2√3)
(42, 2√3)
Solution:
Let the equation the ellipse from equation is:
x2a2 + y2b2 = 1
In an ellipse, the general length of the latus rectum is:
=>2b2a
The general distance between the foci in an ellipse is:
=>2ae
The general length of the minor axis of an ellipse is:
=>2b
From the question,
2b2a = 8 and 2ae = 2b
Now,
=>b(ba) = 4 and ba = e
=>b(e) = 4
Therefore, b = 4/e -(1)
Now, we know that
=>b2 = a2 (1-e2)
=> b2a2 = (1-e2)
Since, b/a = e => e2 = 1 - e2
=>2e2 = 1
Therefore, e = 1√2 ---(2)
From equations (1) and (2), we get
So, b = 4√2
Now,
a2 = b2(1-e2)
=>a2 = 32(1-½)
Therefore, a2 = 64
Now, the ellipse equation is:
x2(64)+y2(32) = 1
Now, we need to find the point that satisfies the obtained equation.
Option (b) (43, 2√2) satisfies the equation. Therefore (b) is the correct answer.
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